Question

In the below-shown figure if $AP=BP$ and $AQ=BQ$, then $AM$ should be equal to $BM$.

• A. True
• B. False

Answer 1

The correct option is A

True

In the given figure,
$PA=PB$ (Given)
$AQ=BQ$ (Given)
$PQ$ is common.
So, $\Delta PAQ$ $\cong$ $\Delta PBQ$ (SSS rule).
$\Rightarrow \angle APQ$=$\angle BPQ$(CPCT).
Also, $PA=PB$ (Given) and $PM$ is common
Hence, $△PMA$ $\cong$ $△PMB$ (SAS Rule)

So, $AM=BM$ and $\angle PMA$ = $\angle PMB$ (CPCT)

$\angle PMA$ + $\angle PMB$ = ${180}^{\circ }$ (Linear pair axiom),

$\angle PMA$ = $\angle PMB$ = ${90}^{\circ }$

Therefore, $PM$, that is, $PMQ$ is the perpendicular bisector of $AB$.

Hence $AM=BM$